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प्रश्न
Fig., below shows the reading obtained while measuring the diameter of a wire with a screw gauge. The screw advances by 1 division on the main scale when circular head is rotated once.
Find:
(i) Pitch of the screw gauge,
(ii) Least count of the screw gauge and
(iii) The diameter of the wire.
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उत्तर
No. of divisions on the circular scale = 50
(i) Pitch = Distance moved ahead in one revolution
= 1 mm/1 = 1 mm.
(ii) L.C. = Pitch/No. of divisions on the circular head
= (1/50) mm
= 0.02 mm
(iii) Main scale reading = 4 mm
No. of circular division coinciding with m.s.d. (p) = 47
Circular scale reading = p × L.C.
= (47 × 0.02) mm
= 0.94 mm
Diameter (Total reading) = M.s.r. + circular scale reading
= (4 + 0.94) mm
= 4.94 mm
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संबंधित प्रश्न
The pitch of a screw gauge is 1 mm and the circular scale has 100 divisions. In measurement of the diameter of a wire, the main scale reads 2 mm and 45th mark on the circular scale coincides with the base line. Find
(i) The least count and
(ii) The diameter of the wire.
How can the least count of a screw gauge be decreased?
In a vernier callipers, 19 main scale divisions coincide with 20 vernier scale divisions. If the main scale has 20 divisions in a centimetre, calculate
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- L.C. of vernier callipers.
In figure for vernier callipers, calculate the length recorded.
Explain the term least count of a screw gauge. How are they determined?
State whether true or false. If false, correct the statement.
With the help of vernier caliper we can have an accuracy of 0.1 mm and with screw gauge we can have an accuracy of 0.01 mm.
On a vernier calliper, 25 vernier scale divisions are equal in length to 24 main scale divisions. One main scale division is 1 mm. Find the least count of the instrument.
In a vernier calipers, (N + 1) divisions of vernier scale coincide with N divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is ______.
